This notion, though probably useful in some ways, has a grave defect. Let
$$ A \xrightarrow{f} B \xrightarrow{g} C, $$
be an exact sequence of groups, so
$$ \mathrm{ker} g = \{y \in B : g(y) = 1\} = \mathrm{img} f = \{f(x) : x \in A\}. $$
I would like to verify that the sequence is exact as a sequence of monoids, so $I_f=K_g$. We have $(y,z) \in K_g$ if and only if $g(yz^{-1}) = 1$ since g is a group homomorphism, if and only if $yz^{-1} \in K = \mathrm{ker} g$ if and only if $y = kz$ for some $k \in K$. Thus
$$ K_g = \{(kz,z) \in B \times B : k \in K, z \in B \}. $$

On the other hand, suppose g is not injective and that f is not surjective (not an
atypical situation). Then there exists $z \in B$ such that $z \not\in f(A)$
and there exists $k \in K$ such that $k \neq 1$. Then $(kz,z) \in K_g$ by
definition, but $(kz,z) \not\in I_f$: it does not belong to $f(A) \times f(A)$
since $z \not \in f(A)$ and does not belong to $\Delta(B)$ since $kz \neq z$. So
$K_g \not\subseteq I_f$.

I can see why one must add the diagonal to $f(A) \times f(A)$, since one
wants a reflexive relation on B. It follows that $I_f$ is an
equivalence relation. I would think that one should go farther and
take the image to be the congruence closure of $I_f$, i.e., the submonoid
$\overline{I}_f$ of $B \times B$ generated by $f(A) \times f(A)$ and $\Delta(B)$. Then it is clear that $(kz,z) = (k,1)(z,z)$ in $\overline{I}_f$, and so
$K_g \subseteq \overline{I}_f$. Conversely, it is clear that $\overline{I}_f$ is
a subset of $K_g$, so then I'm happy again.

But this definition has a different problem: an exact sequence
$$ A \xrightarrow{f} B \to 1$$
of monoids is exact if $f$ is surjective but not conversely: the inclusion map
$$ \mathbb{N} \xrightarrow{f} \mathbb{Z} $$
of the natural numbers in the integers has $\overline{I}_f = \mathbb{Z} \times \mathbb{Z}$ but is not surjective. Arturo Magidin pointed out to me in an e-mail that this map $f$ is an epimorphism in the category of monoids; maybe I'll have to accept that.

Can anyone shed some light on this matter? I'd already be pretty happy with a reference.

To add to Benjamin's and James's responses, any category that is algebraic (or monadic) over $Set$ is Barr-exact.
–
Todd Trimble♦Dec 9 '11 at 23:50

I am always wary of the word 'right' in the first line of the question. "Horses for courses", but given that there was a very nice preprint by Charlie Wells: cwru.edu/artsci/math/wells/pub/pdf/catext.pdf which might be of use. The original dated to 1980.
–
Tim PorterDec 10 '11 at 7:00

2 Answers
2

Hi John. I'd say there is no generalization of short exact sequence to the category of monoids, although I suppose it really depends on what you want to do with it. What you probably want is an internal equivalence relation. So you could say a diagram $A\rightrightarrows B \to C$ of monoid maps (where the two compositions $A\rightrightarrows C$ agree) is short exact if the map $B\to C$ is surjective and the induced map $A\to B\times_C B$ is an isomorphism. This is equivalent to requiring the induced map $A\to B\times B$ to be injective, its image to be an equivalence relation, and the induced map $B/A\to C$ is to be an isomorphism.

My general feeling is that this is the right concept in most categories of sets with algebraic structure (e.g. the category of sets itself, semi-rings). It's only in categories where the objects have some group structure that you can re-express it using kernels.

For the special case of commutative monoids, or more generally semimodules over a semiring, my related preprints on arXiv might be helpful (see below). For arbitrary monoids, the general definition is similar, however it is difficult to apply since the notion of the cokernel of a morphism of monoids is really "slippery":